A153679 - OEIS

%I #15 May 16 2020 14:14:52

%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,

%T 27,28,29,82,134,1306,2036,6393,34477,145984,2746739,2792428,8460321

%N Minimal exponents m such that the fractional part of (1024/1000)^m obtains a maximum (when starting with m=1).

%C Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (1024/1000)^m is greater than the fractional part of (1024/1000)^k for all k, 1<=k<m.

%C The next such number must be greater than 5*10^5.

%C a(40) > 10^7. _Robert Price_, Mar 16 2019

%F Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) > fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).

%e a(30)=82, since fract((1024/1000)^82)= 0.99191990..., but fract((1024/1000)^k)<0.9893 for 1<=k<=81; thus fract((1024/1000)^82)>fract((1024/1000)^k) for 1<=k<82 and 82 is the minimal exponent >29 with this property.

%t $MaxExtraPrecision = 10000;

%t p = 0;

%t Select[Range[1, 50000],

%t If[FractionalPart[(1024/1000)^#] > p,

%t p = FractionalPart[(1024/1000)^#]; True] &] (* _Robert Price_, Mar 16 2019 *)

%o (Python)

%o A153679_list, m, n, k, q = [], 1, 1024, 1000, 0

%o while m < 10**4:

%o     r = n % k

%o     if r > q:

%o         q = r

%o         A153679_list.append(m)

%o     m += 1

%o     n *= 1024

%o     k *= 1000

%o     q *= 1000 # _Chai Wah Wu_, May 16 2020

%Y Cf. A153663, A153671, A153675, A154130, A153687, A153695, A091560, A153711, A153719.

%K nonn,more

%O 1,2

%A _Hieronymus Fischer_, Jan 06 2009

%E a(37)-a(39) from _Robert Price_, Mar 16 2019